LeetCode - Easy - 290. Word Pattern

Topic

• Hash Table

Description

https://leetcode.com/problems/word-pattern/

Given a pattern and a string s, find if s follows the same pattern.

Here follow means a full match, such that there is a bijection between a letter in pattern and a non-empty word in s.

Example 1:

Input: pattern = "abba", s = "dog cat cat dog"
Output: true

Example 2:

Input: pattern = "abba", s = "dog cat cat fish"
Output: false

Example 3:

Input: pattern = "aaaa", s = "dog cat cat dog"
Output: false

Example 4:

Input: pattern = "abba", s = "dog dog dog dog"
Output: false

Constraints:

• 1 <= pattern.length <= 300
• pattern contains only lower-case English letters.
• 1 <= s.length <= 3000
• s contains only lower-case English letters and spaces ' '.
• s does not contain any leading or trailing spaces.
• All the words in s are separated by a single space.

Analysis

双向映射。

注意，Java包装整型的比较问题。

Submission

import java.util.HashMap;
import java.util.Objects;

public class WordPattern {
    
    
	// 方法一：我写的，双向映射
	public boolean wordPattern1(String pattern, String s) {
    
    
		// 双向映射
		HashMap<Character, Integer> aMap = new HashMap<>();
		HashMap<String, Integer> bMap = new HashMap<>();

		String[] words = s.split(" ");

		if (pattern.length() != words.length)
			return false;

		int count = 0;// 这可省略，改用下面循环体的i变量

		for (int i = 0; i < pattern.length(); i++) {
    
    
			Character cr = pattern.charAt(i);
			Integer id1 = aMap.get(cr);
			Integer id2 = bMap.get(words[i]);
			if (id1 == null && id2 == null) {
    
    
				aMap.put(cr, count);
				bMap.put(words[i], count++);
			} else {
    
    
				if (id1 != id2)// 这要改成Objects.equals()
					return false;
			}
		}

		return true;
	}

	// 方法二：双向映射，比方法一更简洁
	public boolean wordPattern2(String pattern, String s) {
    
    
		String[] words = s.split(" ");
		if (words.length != pattern.length())
			return false;
		HashMap<Object, Object> index = new HashMap<>();//Object类型key即可放Character，亦可放String，我没想到
		for (Integer i = 0; i < words.length; ++i)
			if (!Objects.equals(index.put(pattern.charAt(i), i), //
					index.put(words[i], i)))
				return false;
		return true;
	}

}

Test

import static org.junit.Assert.*;

import java.util.Objects;

import org.junit.Test;

public class WordPatternTest {
    
    

	@Test
	public void test() {
    
    
		WordPattern obj = new WordPattern();

		assertTrue(obj.wordPattern1("abba", "dog cat cat dog"));
		assertFalse(obj.wordPattern1("abba", "dog cat cat fish"));
		assertFalse(obj.wordPattern1("aaaa", "dog cat cat dog"));
		assertFalse(obj.wordPattern1("abba", "dog dog dog dog"));
		
		assertTrue(obj.wordPattern2("abba", "dog cat cat dog"));
		assertFalse(obj.wordPattern2("abba", "dog cat cat fish"));
		assertFalse(obj.wordPattern2("aaaa", "dog cat cat dog"));
		assertFalse(obj.wordPattern2("abba", "dog dog dog dog"));
	}
	
	@Test
	public void testEquals() {
    
    
		//== 与 !=运用在Integer包装类的比较的局限性
		//包装类最好用Objects.equals()
		int i = 127;
		Integer a = i;
		Integer b = i;
		assertTrue(a == b);
		assertTrue(Objects.equals(a, b));
		
		i = 128;
		a = i;
		b = i;
		assertFalse(a == b);
		assertTrue(Objects.equals(a, b));
	}
	
}